Danny is trying to figure out how fast sugar crystals can dissolve in water. He has already tested crystals of table sugar by just pouring packets of sugar into a bowl of room-temperature water.
Which should Danny try next if he wants to make the sugar dissolve slower?
A.
stir the water
B.
use larger crystals
C.
heat the water
D.
all of these
Answers: 3
Chemistry, 21.06.2019 20:30, lemonsalt9378
14. complete and balance the equations for the single displacement reactions. a. zn + pb(no3)2 -> b. al + niso4 -> 15. complete and balance the equations for the double displacement reactions. a. agno3(aq) + nacl(aq) -> b. mg(no3)2(aq) + koh(aq) -> 16. complete and balance the equations for the combustion reactions. a. __ ch4 + o2 -> b. __ c3h6 + o2 -> c. + o2 ->
Answers: 2
Chemistry, 22.06.2019 11:00, daniel1480
Problem page combustion of hydrocarbons such as pentane ( c5 h12 ) produces carbon dioxide, a "greenhouse gas." greenhouse gases in the earth's atmosphere can trap the sun's heat, raising the average temperature of the earth. for this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.(a) write a balanced chemical equation, including physical state symbols, for the combustion of liquid pentane into gaseous carbon dioxide and gaseous water. (b) suppose 0.350 kg of pentane are burned in air at a pressure of exactly 1 atm and a temperature of 20.0 degree c. calculate the volume of carbon dioxide gas that is produced. be sure your answer has the correct number of significant digits.
Answers: 2
Chemistry, 22.06.2019 21:50, BookandScienceNerd
Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes
Answers: 1
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